WEBVTT
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Operations on Power Series
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In this video, we’re going to discuss the various different types of power series which can arise from adding, subtracting, and multiplying to power series.
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We’ll discuss finding the radius of convergence of a linear combination of power series.
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We’ll also discuss how to find a power series representation of a function as a combination of power series.
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Finally, we’ll discuss what happens when we multiply two power series together.
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To answer these questions about power series, let’s consider what we already know about regular series.
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If we have two series, the sum over 𝑛 of 𝑎 𝑛 and the sum over 𝑛 of 𝑏 𝑛, which converge and we have a constant 𝑐, then we know the sum over 𝑛 of 𝑎 𝑛 plus the sum over 𝑛 of 𝑏 𝑛 is equal to the sum over 𝑛 of 𝑎 𝑛 plus 𝑏 𝑛.
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And we also know that this will converge.
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Similarly, since 𝑐 is a constant, we know when we’re summing over 𝑛 𝑐 multiplied by 𝑎 𝑛, we can take the constant of 𝑐 outside our sum.
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This gives us 𝑐 multiplied by the sum over 𝑛 of 𝑎 𝑛.
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And this also converges.
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Now, we want to ask the question, what would’ve happened if instead of being given regular series, we were given power series?
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We recall when we’re discussing power series, they don’t converge or diverge.
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Instead, they have what we call a radius of convergence.
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This is because 𝑥 is a variable.
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So for each value of 𝑥, we’ll have a different power series.
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And we want to know the values of 𝑥 for which these series converges.
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So let’s add radius of convergences for both of our power series.
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Let’s call these 𝑅 one and 𝑅 two.
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We now want to know what happens when we try to add our power series together.
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We’ll do this by using our formula for adding two series together.
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We’re trying to add two series together.
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The summand in the first one is 𝑎 𝑛 multiplied by 𝑥 to the 𝑛th power.
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And the summand of the second one is 𝑏 𝑛 multiplied by 𝑥 to the 𝑛th power.
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And our formula tells us that when these series converges, we can just add the summands together.
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So for any value of 𝑥 which both of our power series converge, we can use this formula.
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This gives us the sum over 𝑎 𝑛 of 𝑎 𝑛 multiplied by 𝑥 to the 𝑛th power plus 𝑏 𝑛 multiplied by 𝑥 to the 𝑛th power.
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And we can factor out 𝑥 to the 𝑛th power.
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This gives us the sum over [𝑛] of 𝑎 𝑛 plus 𝑏 𝑛 multiplied by 𝑥 to the 𝑛th power.
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We notice, in particular, that this new formula will always work when the absolute value of 𝑥 is less than the smaller of 𝑅 one and 𝑅 two.
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This is because if the absolute value of 𝑥 is less than the smaller of 𝑅 one and 𝑅 two, then it’s smaller than both of them.
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So both of our power series will converge.
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We might be tempted at this point to call the smaller of 𝑅 one and 𝑅 two the radius of convergence for the sum of our two power series.
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However, this is not necessarily the case.
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To see why finding the radius of convergence of this series is not so simple, consider the following example.
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The sum over 𝑛 of 𝑥 to the 𝑛th power divided by 𝑛 will converge when the absolute value of 𝑥 is less than one.
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Similarly, the sum over 𝑛 of negative 𝑥 to the 𝑛th power of 𝑛 will also converge when the absolute value of 𝑥 is less than one.
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However, adding these two power series together gives us the sum of one over 𝑛 minus one over 𝑛 multiplied by 𝑥 to the 𝑛th power.
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Which simplifies to just give us the sum over 𝑛 of zero, which converges for all values of 𝑥.
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So we can use this formula to add any two power series together.
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However, we’re going to have to work hard to find the radius of convergence of our new power series.
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We’ll have to use techniques such as inspection, the ratio test, or the integral test to find our new radius of convergence.
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Let’s take a look at an example.
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Suppose the sum from 𝑛 equals zero to ∞ of 𝑎 𝑛 multiplied by 𝑥 to the 𝑛th power is a power series whose interval of convergence is the open interval from negative three to three.
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And the sum from 𝑛 equals zero to ∞ of 𝑏 𝑛 multiplied by 𝑥 to the 𝑛th power is a power series whose interval of convergence is the open interval from negative five to five.
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Find the interval of convergence of the series the sum from 𝑛 equals zero to ∞ of 𝑎 𝑛 𝑥 to the 𝑛th power minus 𝑏 𝑛 𝑥 to the 𝑛th power.
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And find the [interval] of convergence of the series the sum from 𝑛 equals zero to ∞ of 𝑏 𝑛 two the 𝑛th power 𝑥 to the 𝑛th power.
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The question wants us to find the interval of convergence for two different power series.
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We recall that an interval of convergence is an interval which contains all values of 𝑥 such that our series converges.
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If both power series converges so the absolute value of 𝑥 is less than the smaller of three or five.
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Therefore, since we’re adding together two convergent power series, we can combine the summands just as we would in a regular series.
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This gives us the sum from 𝑛 equals zero to ∞ of 𝑎 𝑛 𝑥 to the 𝑛th power minus 𝑏 𝑛 𝑥 to the 𝑛th power will converge when the absolute value of 𝑥 is less than the smaller of three and five.
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So we’ve shown that the power series converges when the absolute value of 𝑥 is less than three.
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But what about when it’s equal to three?
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What about when it’s greater than three?
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Let’s consider the case when the absolute value of 𝑥 is equal to three.
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And let’s assume that our series converges.
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Then, since the absolute value of 𝑥 is equal to three, the sum from 𝑛 equals zero to ∞ of 𝑏 𝑛 𝑥 to the 𝑛th power must also converge because its radius of convergence is five.
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Now, what would happen if we tried to add these two series together?
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Well, we’ve assumed that our first power series will converge when the absolute value of 𝑥 is equal to three.
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And if the absolute value of 𝑥 is equal to three, this is less than five.
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So our second power series must also converge.
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This means we’re trying to add together two power series which both converge.
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So we can just add the coefficients of 𝑥 to the 𝑛th power together.
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We see that negative 𝑏 𝑛 and 𝑏 𝑛 cancel.
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So this simplifies to give us the sum from 𝑛 equals zero to ∞ of 𝑎 𝑛 multiplied by 𝑥 to the 𝑛th power.
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However, the sum from 𝑛 equals zero to ∞ of 𝑎 𝑛 𝑥 to the 𝑛th power will only converge when the absolute value of 𝑥 is less than three.
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So this series must diverge.
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The only way this could’ve happened is if the sum over 𝑛 of 𝑎 𝑛 𝑥 to the 𝑛th power minus 𝑏 𝑛 𝑥 to the 𝑏 𝑛 𝑥 to the 𝑛th power was divergent when the absolute value of 𝑥 was equal to three.
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Therefore, since neither three nor negative three are in our interval of convergence, we’ve shown that the sum from 𝑛 equals zero to ∞ of 𝑎 𝑛 𝑥 to the 𝑛th power minus 𝑏 𝑛 𝑥 to the 𝑛th power has the interval of convergence the open interval from negative three to three.
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So let’s clear some space and work on the second case of finding our interval of convergence.
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We’ll let the function 𝑓 of 𝑥 be equal to the sum from 𝑛 equals zero to ∞ of 𝑏 𝑛 𝑥 to the 𝑛th power.
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And this will be true whenever the absolute value of 𝑥 is less than five.
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We want to make this power series look like the one given to us in the question.
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We’ll start by multiplying and dividing our summands by two to the 𝑛th power.
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Next, we can use our laws of exponents to notice that a half to the 𝑛th power multiplied by 𝑥 to the 𝑛th power is equal to 𝑥 over two all raised to the 𝑛th power.
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We want our summands to have 𝑥 to the 𝑛th power instead of 𝑥 over two to the 𝑛th power.
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So we’ll let 𝑢 be equal to 𝑥 divided by two.
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This gives us that the sum from 𝑛 equals zero to ∞ of 𝑏 𝑛 two to the 𝑛th power multiplied by 𝑢 to the 𝑛th power will converge when the absolute value of two 𝑢 is less than five.
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If the absolute value of two 𝑢 is less than five, this is the same as saying the absolute value of 𝑢 is less than five over two.
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So we’ve shown that the radius of convergence of the sum from 𝑛 equals zero to ∞ of 𝑏 𝑛 multiplied by two to the 𝑛th power multiplied by 𝑢 to the 𝑛th power is five over two.
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Since it doesn’t matter if we call our variable 𝑢 or 𝑥, we’ve shown that our second power series, the sum from 𝑛 equals zero to ∞ of 𝑏 𝑛 two to the 𝑛th power 𝑥 to the 𝑛th power, will have the interval of convergence the open interval from negative five over two to five over two.
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Let’s now take a look at what would happen if we tried to multiply two power series together.
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So let’s say we wanted to multiply two power series together, how would we do this?
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Well, let’s start by writing each out by term by term.
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Writing these out term by term gives us the following expression.
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Well, we remember that both of these expressions go on infinitely.
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If we look at the two expressions we have, these just seem like two polynomials.
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And we know how to multiply two polynomials together.
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To find the constant term, we multiply both of our constant terms together.
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This gives us 𝑎 nought multiplied by 𝑏 nought.
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Now, to find all our terms of 𝑥, we multiply the constant terms by the coefficients with 𝑥.
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This gives us 𝑎 nought multiplied by 𝑏 one plus 𝑎 one multiplied by 𝑏 nought.
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We can then do the same for the coefficient of 𝑥 squared.
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To get our terms of 𝑥 squared in the product, we’ll be multiplying the constant terms by the terms of 𝑥 squared.
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And we’ll also be multiplying the terms of a single 𝑥 in them together.
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And we could continue doing this process indefinitely.
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We can now notice when we were choosing our coefficient to multiply to get the term 𝑥 squared in our product, we were choosing them so that when we sum their indexes, we get two.
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And this makes sense because the index of our coefficient is exactly the same as the power of 𝑥.
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Let’s now take a closer look at our second term.
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We could write this as the sum from 𝑗 equals zero to one of 𝑎 𝑗 multiplied by 𝑏 one minus 𝑗 all multiplied by 𝑥 to the first power.
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And again, this is because we wanted the sum of our indexes to be equal to the power of 𝑥.
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We could do exactly the same for the coefficient of 𝑥 squared.
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In fact, we could make a similar sum as the coefficient of any power of 𝑥.
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We can finally write this as a sum over our different powers of 𝑥.
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Where each coefficient of our power of 𝑥 will be equal to one of the coefficients we found earlier.
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As we discussed earlier, to find the coefficient of 𝑥 to the 𝑛th power, we need to add together the products of all terms whose indexes add to 𝑛.
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And this gives us a motivation of how to multiply two power series together.
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If we’re given two power series which converge for a certain value of 𝑥, then the product of these two power series is equal to.
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We sum over 𝑛 to find the coefficient of 𝑥 to the 𝑛th power.
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And this coefficient is the sum over 𝑗 of 𝑎 𝑗 multiplied by 𝑏 𝑛 minus 𝑗.
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Let’s take a look at an example to see how we can use this formula in practice.
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Multiply the series one divided by one plus 𝑥 is equal to the sum from 𝑛 equals zero to ∞ of negative one to the 𝑛th power 𝑥 to the 𝑛th power by itself to construct a series for one divided by one plus 𝑥 squared.
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Write the answer in sigma notation.
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We recall if we have two power series, 𝑓 of 𝑥 is equal to the sum over 𝑛 of 𝑎 𝑛 𝑥 to the 𝑛th power and 𝑔 of 𝑥 is equal to the sum over 𝑛 of 𝑏 𝑛 𝑥 to the 𝑛th power, which both converge for this value of 𝑥.
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Then we can calculate the product of these two power series, 𝑓 of 𝑥 multiplied by 𝑔 of 𝑥, as the sum over 𝑛 of all of our 𝑥 to the 𝑛th power terms with the coefficient the sum over 𝑗 𝑎 𝑗 multiplied by 𝑏 𝑛 minus 𝑗.
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In our case, we want to multiply one divided by one plus 𝑥 by itself to get the power series for one divided by one plus 𝑥 squared.
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So we set both of our functions, 𝑓 of 𝑥 and 𝑔 of 𝑥, to be equal to one divided by one plus 𝑥.
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So we have both 𝑓 of 𝑥 and 𝑔 of 𝑥 are equal to the sum of one over one plus 𝑥 which are equal to the sum from 𝑛 equals zero to ∞ of negative one to the 𝑛th power 𝑥 to the 𝑛th power.
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So for the value of 𝑥 where our power series for 𝑓 of 𝑥 converges, we can use our formula to find an expression for 𝑓 of 𝑥 squared.
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The question tells us that one divided by one plus 𝑥 is equal to the sum from 𝑛 equals zero to ∞ of negative one to the 𝑛th power 𝑥 to the 𝑛th power.
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This gives us that 𝑎 𝑗 is negative one raised to the power of 𝑗.
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And 𝑏 𝑛 minus 𝑗 is negative one raised to the power of 𝑛 minus 𝑗.
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Therefore, we can calculate 𝑎 𝑗 multiplied by 𝑏 𝑛 minus 𝑗 as negative one to the power of 𝑗 multiplied by negative one to the power of 𝑛 minus 𝑗, which is just negative one to the 𝑛th power.
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So we now have our coefficient of 𝑥 to the power of 𝑛 is the sum from 𝑗 equals zero to 𝑛 of negative one to the 𝑛th power.
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And we can see that’s our summand, negative one to the 𝑛th power, is independent of 𝑗.
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So, in fact, our sum is negative one to the 𝑛th power added to itself 𝑛 plus one times.
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This is because 𝑗 goes from zero to 𝑛.
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And if we’re adding negative one to the 𝑛th power to itself 𝑛 plus one times, this is the same as saying 𝑛 plus one multiplied by negative one to the 𝑛th power.
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Therefore, we’ve shown for the values of 𝑥 where a power series for one divided by one plus 𝑥 converges.
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We can multiply the power series of one divided by one plus 𝑥 by itself to get that one divided by one plus 𝑥 squared is equal to the sum from 𝑛 equals zero to ∞ of negative one to the 𝑛th power multiplied by 𝑛 plus one multiplied by 𝑥 to the 𝑛th power.
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So to summarize what we’ve shown in this video, if we have two power series, the sum from 𝑛 equals zero to ∞ of 𝑎 𝑛 𝑥 to the 𝑛th power and the sum from 𝑛 equals zero to ∞ of 𝑏 𝑛 𝑥 to the 𝑛th power, which both converge for a particular value of 𝑥.
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Then we can add these two power series together to give us the sum from 𝑛 equals zero to ∞ of 𝑎 𝑛 plus 𝑏 𝑛 all multiplied by 𝑥 to the 𝑛th power.
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And, in particular, if our first power series has a radius of convergence 𝑅 one and our second power series has a radius of convergence 𝑅 two.
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Then we know we can add the power series in this way for any value of 𝑥 such that the absolute value of 𝑥 is less than the smaller of 𝑅 one and 𝑅 two.
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And for a power series, the sum from 𝑛 equals zero to ∞ of 𝑎 𝑛 plus 𝑏 𝑛 all multiplied by 𝑥 to the 𝑛th power, if we call its radius of convergence 𝑅, then the minimum of 𝑅 one and 𝑅 two gives us a lower bound on 𝑅.
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Finally, we were shown if we were given two power series, the sum from 𝑛 equals zero to ∞ of 𝑎 𝑛 𝑥 to the 𝑛th power and the sum from 𝑛 equals zero to ∞ of 𝑏 𝑛 𝑥 to the 𝑛th power, which both converge for a value of 𝑥.
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Then the product of these two power series is given by.
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We sum over all of our 𝑥 to the 𝑛th terms.
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And their coefficient is the sum from 𝑗 equals zero to 𝑛 of 𝑎 𝑗 multiplied by 𝑏 𝑛 minus 𝑗.